Explanation

(b)
The general term in the expansion of $\left( \sqrt{\frac{x}{3}} + \frac{3}{{2x}^{2}} \right)^{10}$ is
$T_{r + 1} = \ ^{10}C_{r}\left( \frac{x}{3} \right)^{\frac{10 - r}{2}}\left( \frac{3}{2x^{2}} \right)^{r}$
$= \ ^{10}C_{r}3^{\frac{- 10 + 3r}{2}}.2^{- r}.x^{\frac{10 - 5r}{2}}$
For independent of $x$,
$\frac{10 - 5r}{2} = 0\ \Rightarrow \ r = 2$
$\therefore\ T_{3} = \ ^{10}C_{2} \times \left( \frac{1}{3} \right)^{4}\left( \frac{3}{2} \right)^{2}$
$= \frac{10 \times 9}{2 \times 1} \times \frac{1}{3 \times 3 \times 2 \times 2} = \frac{5}{4}$